sicp/2-3.org

#+BEGIN_HTML
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title: 2.3 - Symbolic Data
layout: org
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* Quotation
** Exercise 2.53
   What would the interpreter print in response to evaluating each of
   the following expressions?

   #+BEGIN_SRC scheme
     (list 'a 'b 'c)

     (list (list 'george))

     (cdr '((x1 x2) (y1 y2)))

     (cadr '((x1 x2) (y1 y2)))

     (pair? (car '(a short list)))

     (memq 'red '((red shoes) (blue socks)))

     (memq 'red '(red shoes blue socks))
   #+END_SRC

   ----------------------------------------------------------------------

   #+BEGIN_SRC scheme :tangle yes
     ;; -------------------------------------------------------------------
     ;; Exercise 2.53
     ;; -------------------------------------------------------------------

     (list 'a 'b 'c)
     ;; (a b c)
     (list (list 'george))
     ;; ((george))
     (cdr '((x1 x2) (y1 y2)))
     ;; ((y1 y2))
     (cadr '((x1 x2) (y1 y2)))
     ;; (y1 y2)
     (pair? (car '(a short list)))
     ;; #f
     (memq 'red '((red shoes) (blue socks)))
     ;; #f
     (memq 'red '(red shoes blue socks))
     ;; (red shoes blue socks)
   #+END_SRC
** Exercise 2.54
   Two lists are said to be `equal?' if they contain
   equal elements arranged in the same order.  For example,

   #+BEGIN_SRC scheme
          (equal? '(this is a list) '(this is a list))
   #+END_SRC
   
   is true, but

   #+BEGIN_SRC scheme
   (equal? '(this is a list) '(this (is a) list))
   #+END_SRC

   is false.  To be more precise, we can define `equal?'  recursively
   in terms of the basic `eq?' equality of symbols by saying that `a'
   and `b' are `equal?' if they are both symbols and the symbols are
   `eq?', or if they are both lists such that `(car a)' is `equal?'
   to `(car b)' and `(cdr a)' is `equal?' to `(cdr b)'.  Using this
   idea, implement `equal?' as a procedure.(5)

   ----------------------------------------------------------------------

   #+BEGIN_SRC scheme :tangle yes
     ;; -------------------------------------------------------------------
     ;; Exercise 2.54
     ;; -------------------------------------------------------------------

     (define (my-equal? a b)
       (cond ((and (null? a)
                   (null? b))
              #t)
             ((and (symbol? a)
                   (symbol? b))
              (eq? a b))
             ((and (pair? a)
                   (pair? b))
              (and (eq? (car a) (car b))
                   (my-equal? (cdr a) (cdr b))))
             (else #f)))
   #+END_SRC
   
** Exercise 2.55
   Eva Lu Ator types to the interpreter the expression

   #+BEGIN_SRC scheme
     (car ''abracadabra)
   #+END_SRC

   To her surprise, the interpreter prints back `quote'.  Explain.

   ----------------------------------------------------------------------

   ~'abracadabra~ expands to ~(quote abracadabra)~ Therefore
   ~''abracadabra~ expands to ~'(quote abracadabra)~, which expands to
   ~(quote (quote abracadabra))~, the car of which is the atom
   ~quote~.
* Example: Symbolic Differentiation
* Example: Representing Sets
* Example: Huffman Encoding Trees